For all n in N,if 1^3+2^3+3^3+ldots+n^3 x,t | vedclass

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(A) The sum of the cubes of the first $n$ natural numbers is given by the formula: $S_n = 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[ \frac{n(n+1)}{2} \ri

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$\frac{n^2}{4}$

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(A) The sum of the cubes of the first $n$ natural numbers is given by the formula: $S_n = 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$. Given the inequality $1^3 + 2^3 + 3^3 + \ldots + n^3 > x$,we substitute the sum formula: $\frac{n^2(n+1)^2}{4} > x$. Since $\frac{n^2(n+1)^2}{4}$ is the exact sum,any value $x$ that is strictly less than this sum satisfies the condition. Comparing the given options: $A) \frac{n^2}{4} $B) n^2  1$. $C) n^4$ is not necessarily less than $\frac{n^2(n+1)^2}{4}$. $D) \frac{n^2(n+1)^2}{4}$ is equal to the sum,not strictly less than it. Among the choices,$\frac{n^2}{4}$ is a value that is always less than the sum for all $n \in N$.

For all $n \in N$,if $1^3+2^3+3^3+\ldots+n^3 > x$,then a value of $x$ among the following is

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